**
**

The net for this box would be:

The **volume **of the box is *V *=* x*^{2}*y*

We are told that the surface area of the box is 64 cm^{2}. The area of the base of the box is *x*^{2} and the area of each side is *xy*, so the area of the base plus the area of the 4 sides is given by:

x^{2}+ 4xy= 64 cm^{2}

Solving for *y* gives:

`y=(64-x^2)/(4x)=16/x-x/4`

So the volume can be rewritten:

`V=x^2y`

`=x^2(16/x-x/4)`

`=16x-x^3/4`

Now

`(dV)/(dx)=16-(3x^2)/4`

and this is zero when

`x=+- 8/sqrt(3) ~~ 4.62`

(**Note:** The negative case has no practical meaning.)

**Is it a maximum? **

`(d^2V)/(dx^2)=-(3x)/2`

and this is negative when *x* is positive.
So it is a MAX.

So the **dimensions** of the box are:

Base 4.62 cm × 4.62 cm and sides 2.31 cm.

The **maximum possible volume** is

*V* = 4.62 × 4.62 × 2.31 ≈ 49.3 cm^{3}

*Check*: Area of material:

*x*^{2} + 4*xy* = 21.3 + 4 × 4.62 × 2.31 = 64

Checks OK.

Easy to understand math videos:

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